A bulb and a capacitor are connected in serie | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The impedance $Z$ of an $RC$ series circuit is given by $Z = \sqrt{R^2 + X_C^2}$,where $X_C = \frac{1}{2\pi f C}$ is the capacitive reactance.As t

Vedclass · Accepted Answer

Bulb will give more intense light

Vedclass · Answer

(A) The impedance $Z$ of an $RC$ series circuit is given by $Z = \sqrt{R^2 + X_C^2}$,where $X_C = \frac{1}{2\pi f C}$ is the capacitive reactance.As the frequency $f$ of the $ac$ source increases,the capacitive reactance $X_C$ decreases.Since $Z = \sqrt{R^2 + (\frac{1}{2\pi f C})^2}$,a decrease in $X_C$ leads to a decrease in the total impedance $Z$ of the circuit.According to Ohm's law for $ac$ circuits,the current $I = \frac{V}{Z}$. Since $V$ is constant and $Z$ decreases,the current $I$ in the circuit increases.The intensity of light emitted by the bulb is proportional to the power dissipated,$P = I^2 R$. As the current $I$ increases,the power dissipated increases,and the bulb glows more intensely.

$A$ bulb and a capacitor are connected in series to a source of alternating current. If its frequency is increased,while keeping the voltage of the source constant,then

Similar Questions

The correct structure of the drug paracetamol is:

If a body is moving in a circle of radius $r$ with a constant speed $v$,its angular velocity is

$\left|\begin{array}{ccc}a-b-c & 2a & 2a \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b\end{array}\right|$ is equal to

$A$ clock $S$ is based on the oscillation of a spring,and a clock $P$ is based on pendulum motion. Both clocks run at the same rate on Earth. On a planet having the same density as Earth but twice the radius,which of the following is true?

If $A$ and $B$ are any two events,then $P(A \cup B) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module